The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of `9.0xx10^(3)` km. Find the mass of Mars.
`{"Given "(4pi^(2))/(G)=6xx10^(11)N^(-1)m^(-2)kg^(2)}`

Options

A. `5.96xx10^(19)" kg"`
B. `3.25xx10^(21)" kg"`
C. `7.02xx10^(25)" kg"`
D. `6.00xx10^(23)" kg"`

Answer

To find the mass of Mars using the given information about one of its moons, we can follow these steps:

### Step 1: Convert the period from hours and minutes to seconds
The period \( T \) is given as 7 hours and 30 minutes. We need to convert this into seconds.

\[ T = 7 \text{ hours} + 30 \text{ minutes} = (7 \times 3600) + (30 \times 60) \text{ seconds} \]
\[ T = 25200 + 1800 = 27000 \text{ seconds} \]

### Step 2: Use the formula for gravitational force and centripetal force
The centripetal force acting on the moon is provided by the gravitational force exerted by Mars. We can equate these two forces.

\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]

Where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of Mars,
- \( m \) is the mass of the moon,
- \( r \) is the orbital radius,
- \( v \) is the orbital velocity of the moon.

### Step 3: Express the orbital velocity \( v \) in terms of the period \( T \)
The orbital velocity \( v \) can be expressed as:

\[ v = \frac{2 \pi r}{T} \]

### Step 4: Substitute \( v \) into the centripetal force equation
Substituting \( v \) into the centripetal force equation gives:

\[ \frac{G M m}{r^2} = \frac{m}{r} \left(\frac{2 \pi r}{T}\right)^2 \]

### Step 5: Cancel out \( m \) and rearrange the equation
We can cancel \( m \) from both sides (since it is not zero) and rearrange the equation:

\[ \frac{G M}{r^2} = \frac{4 \pi^2 r}{T^2} \]

### Step 6: Solve for the mass \( M \) of Mars
Rearranging the equation to solve for \( M \):

\[ M = \frac{4 \pi^2 r^3}{G T^2} \]

### Step 7: Substitute the known values
We know:
- \( r = 9.0 \times 10^3 \text{ km} = 9.0 \times 10^6 \text{ m} \)
- \( T = 27000 \text{ seconds} \)
- \( \frac{4 \pi^2}{G} = 6 \times 10^{11} \text{ N}^{-1} \text{ m}^{-2} \text{ kg}^{2} \)

Substituting these values into the equation:

\[ M = \left(6 \times 10^{11}\right) \cdot \left(9.0 \times 10^6\right)^3 \cdot \frac{1}{(27000)^2} \]

### Step 8: Calculate \( M \)
Calculating \( (9.0 \times 10^6)^3 \):

\[ (9.0 \times 10^6)^3 = 729 \times 10^{18} = 7.29 \times 10^{20} \]

Now substituting this back into the equation for \( M \):

\[ M = 6 \times 10^{11} \cdot \frac{7.29 \times 10^{20}}{(27000)^2} \]

Calculating \( (27000)^2 \):

\[ (27000)^2 = 729000000 = 7.29 \times 10^8 \]

Now substituting this value:

\[ M = 6 \times 10^{11} \cdot \frac{7.29 \times 10^{20}}{7.29 \times 10^8} \]

This simplifies to:

\[ M = 6 \times 10^{11} \cdot 10^{12} = 6 \times 10^{23} \text{ kg} \]

### Final Answer
The mass of Mars is approximately \( 6 \times 10^{23} \text{ kg} \).

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